Given the differential equation: \[ y'' - 3y' + 4y = 2e^{-x} \] and a known solution to the associated homogeneous equation: \[ y_1(x) = e^{-x} \] We are to **find the general solution** using **reduction of order**. --- ## **Step 1: Find the general solution to the homogeneous equation** The homogeneous equation is: \[ y'' - 3y' + 4y = 0 \] Let \( y = e^{rx} \): \[ r^2 - 3r + 4 = 0 \] Solve the characteristic equation: \[ r = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot 4}}{2} = \frac{3 \pm \sqrt{9 - 16}}{2} = \frac{3 \pm \sqrt{-7}}{2} = \frac{3}{2} \pm \frac{\sqrt{7}}{2}i \] So the general solution to the homogeneous equation is: \[ y_h(x) = e^{\frac{3}{2}x} \left[ C_1 \cos\left(\frac{\sqrt{7}}{2}x\right) + C_2 \sin\left(\frac{\sqrt{7}}{2}x\right) \right] \] But **this doesn't match the given solution \( y_1(x) = e^{-x} \)**. Let's check the calculation: Given \( y_1(x) = e^{-x} \), try plugging in \( r = -1 \): \[ (-1)^2 - 3(-1) + 4 = 1 + 3 + 4 = 8 \neq 0 \] So \( e^{-x} \) is **not a solution to the homogeneous equation** as written. However, the problem states that it is, so let's proceed **with the given function** per the instructions, and use reduction of order. --- ## **Step 2: Use reduction of order** Let \( y_2(x) = v(x) y_1(x) = v(x) e^{-x} \). Compute derivatives: \[ y_2' = v' e^{-x} + v (-e^{-x}) = (v' - v) e^{-x} \] \[ y_2'' = [v'' - v'] e^{-x} - (v' - v) e^{-x} = (v'' - v' - v' + v) e^{-x} = (v'' - 2v' + v) e^{-x} \] Plug into the homogeneous equation: \[ y_2'' - 3y_2' + 4y_2 = 0 \] \[ (v'' - 2v' + v)e^{-x} - 3[(v' - v)e^{-x}] + 4[v e^{-x}] = 0 \] \[ (v'' - 2v' + v - 3v' + 3v + 4v) e^{-x} = 0 \] \[ [v'' - 2v' + v - 3v' + 3v + 4v] = v'' - 5v' + 8v \] So, \[ v'' - 5v' + 8v = 0 \] The characteristic equation: \[ r^2 - 5r + 8 = 0 \implies r = \frac{5 \pm \sqrt{25 - 32}}{2} = \frac{5 \pm i\sqrt{7}}{2} \] So, \[ v(x) = e^{\frac{5}{2}x} [A \cos(\frac{\sqrt{7}}{2}x) + B \sin(\frac{\sqrt{7}}{2}x)] \] Thus, \[ y_2(x) = v(x) e^{-x} = e^{\frac{5}{2}x} e^{-x} [A \cos(\frac{\sqrt{7}}{2}x) + B \sin(\frac{\sqrt{7}}{2}x)] = e^{\frac{3}{2} x}[A \cos(\frac{\sqrt{7}}{2}x) + B \sin(\frac{\sqrt{7}}{2}x)] \] So the general solution to the **homogeneous** equation is: \[ y_h(x) = C_1 e^{-x} + C_2 e^{\frac{3}{2} x} \cos\left(\frac{\sqrt{7}}{2} x\right) + C_3 e^{\frac{3}{2} x} \sin\left(\frac{\sqrt{7}}{2} x\right) \] But only two linearly independent solutions are needed. So, \[ y_h(x) = C_1 e^{-x} + C_2 e^{\frac{3}{2} x} \cos\left(\frac{\sqrt{7}}{2} x\right) + C_3 e^{\frac{3}{2} x} \sin\left(\frac{\sqrt{7}}{2} x\right) \] or usually, \[ y_h(x) = C_1 e^{-x} + C_2 e^{\frac{3}{2} x} \cos\left(\frac{\sqrt{7}}{2} x\right) + C_3 e^{\frac{3}{2} x} \sin\left(\frac{\sqrt{7}}{2} x\right) \] --- ## **Step 3: Find a particular solution using Variation of Parameters** Since the nonhomogeneous term is \( 2e^{-x} \), and \( e^{-x} \) is already a solution to the homogeneous equation, try: Let \( y_p(x) = x e^{-x} \) (method of undetermined coefficients). Compute derivatives: \[ y_p' = e^{-x} - x e^{-x} \] \[ y_p'' = -e^{-x} - [e^{-x} - x e^{-x}] = -e^{-x} - e^{-x} + x e^{-x} = -2e^{-x} + x e^{-x} \] Plug into the equation: \[ y_p'' - 3y_p' + 4y_p = (-2e^{-x} + x e^{-x}) - 3[e^{-x} - x e^{-x}] + 4 x e^{-x} \] \[ = -2e^{-x} + x e^{-x} -3e^{-x} + 3x e^{-x} + 4x e^{-x} \] \[ = (-2e^{-x} - 3e^{-x}) + (x e^{-x} + 3x e^{-x} + 4x e^{-x}) \] \[ = -5e^{-x} + 8x e^{-x} \] We want this to equal \( 2e^{-x} \). It's not a solution; let's try \( y_p(x) = A x e^{-x} \). So, \[ y_p = A x e^{-x} \] \[ y_p' = A e^{-x} - A x e^{-x} \] \[ y_p'' = -A e^{-x} - [A e^{-x} - A x e^{-x}] = -A e^{-x} - A e^{-x} + A x e^{-x} = -2A e^{-x} + A x e^{-x} \] Plug into the equation: \[ y_p'' - 3y_p' + 4y_p = (-2A e^{-x} + A x e^{-x}) - 3[A e^{-x} - A x e^{-x}] + 4A x e^{-x} \] \[ = -2A e^{-x} + A x e^{-x} - 3A e^{-x} + 3A x e^{-x} + 4A x e^{-x} \] \[ = (-2A e^{-x} - 3A e^{-x}) + (A x e^{-x} + 3A x e^{-x} + 4A x e^{-x}) \] \[ = -5A e^{-x} + 8A x e^{-x} \] Set equal to \( 2e^{-x} \): \[ -5A e^{-x} + 8A x e^{-x} = 2 e^{-x} \] For all \( x \), coefficients must be equal: - For \( x e^{-x} \): \( 8A = 0 \implies A = 0 \) - For \( e^{-x} \): \( -5A = 2 \implies A = -\frac{2}{5} \) But that would contradict the other result. This means **method of undetermined coefficients does not work**; try **variation of parameters**. ### **Variation of Parameters** We have two linearly independent solutions: - \( y_1(x) = e^{-x} \) - \( y_2(x) = e^{\frac{3}{2} x} \cos\left( \frac{\sqrt{7}}{2} x \right) \) Let \( y_p = u_1(x) y_1(x) + u_2(x) y_2(x) \). But since the nonhomogeneous term matches \( y_1 \), the particular solution has the form \( y_p(x) = x e^{-x} \cdot A \). We already showed above that this gives \( -5A e^{-x} + 8A x e^{-x} = 2 e^{-x} \), which cannot be satisfied for all \( x \). So the correct approach is **variation of parameters**. Let the particular solution be: \[ y_p = v_1(x) y_1(x) + v_2(x) y_2(x) \] Let’s use the formula: \[ y_p = -y_1(x) \int \frac{y_2(x) g(x)}{W(y_1, y_2)} dx + y_2(x) \int \frac{y_1(x) g(x)}{W(y_1, y_2)} dx \] Where \( g(x) = 2 e^{-x} \). Let’s choose \( y_1(x) = e^{-x} \), \( y_2(x) = e^{\frac{3}{2} x} \cos\left(\frac{\sqrt{7}}{2} x\right) \). Compute the Wronskian: \[ W(y_1, y_2) = \begin{vmatrix} e^{-x} & e^{\frac{3}{2} x} \cos\left(\frac{\sqrt{7}}{2} x\right) \\ -e^{-x} & \frac{d}{dx} e^{\frac{3}{2} x} \cos\left(\frac{\sqrt{7}}{2} x\right) \end{vmatrix} \] First, compute \( y_2'(x) \): \[ y_2(x) = e^{\frac{3}{2} x} \cos\left( \frac{\sqrt{7}}{2} x \right ) \] \[ y_2'(x) = e^{\frac{3}{2} x} \left( \frac{3}{2} \cos\left( \frac{\sqrt{7}}{2} x \right ) - \frac{\sqrt{7}}{2} \sin\left( \frac{\sqrt{7}}{2} x \right ) \right ) \] So, \[ W = e^{-x} \cdot e^{\frac{3}{2} x} \left( \frac{3}{2} \cos\left( \frac{\sqrt{7}}{2} x \right ) - \frac{\sqrt{7}}{2} \sin\left( \frac{\sqrt{7}}{2} x \right ) \right ) - [e^{\frac{3}{2} x} \cos\left(\frac{\sqrt{7}}{2} x\right)] \cdot [ -e^{-x} ] \] \[ = e^{-x} e^{\frac{3}{2} x} \left( \frac{3}{2} \cos(\cdots) - \frac{\sqrt{7}}{2} \sin(\cdots) \right ) + e^{-x} e^{\frac{3}{2} x} \cos(\cdots) \] \[ = e^{-x} e^{\frac{3}{2} x} \left( \frac{3}{2} \cos(\cdots) - \frac{\sqrt{7}}{2} \sin(\cdots) + \cos(\cdots) \right ) \] \[ = e^{\frac{1}{2} x} \left( \frac{5}{2} \cos(\cdots) - \frac{\sqrt{7}}{2} \sin(\cdots) \right ) \] Let’s denote \( \theta = \frac{\sqrt{7}}{2} x \): \[ W = e^{\frac{1}{2} x} \left( \frac{5}{2} \cos\theta - \frac{\sqrt{7}}{2} \sin\theta \right ) \] Now, plug into the variation